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MATEMATICKI VESNIK
MATEMATIQKI VESNIK
72, 1 (2020), 17–29
March 2020
research paper
originalni nauqni rad
AN EXISTENCE THEOREM OF TRIPLED FIXED POINT FOR ACLASS OF OPERATORS ON BANACH SPACE WITH
APPLICATIONS
Behnam Matani and Jamal Rezaei Roshan
Abstract. In this paper, using the technique of measure of noncompactness, we provesome theorems on tripled fixed points for a class of operators in a Banach space. Also as anapplication, we discuss the existence of solution for a class of systems of nonlinear functionalintegral equations. Finally a concrete example illustrating the mentioned applicability is alsoincluded.
1. Introduction
The study of nonlinear integral equations is a subject of interest for researchers inthe nonlinear functional analysis. Integral equations occur in many applications, inapplied mathematics, as well as in physics. In this context, several authors havepresented papers on the existence of solution for such equations. On the other hand,measure of noncompactness and Darbo’s fixed point theorem are two main tools forproving this result. So far different definitions of measure of noncompactness havebeen suggested by many authors [7, 8, 16]. In this paper, we accept the definitionwhich is presented in [8] and is convenient in application. Up to now, several authorshave presented papers on the existence of solution for nonlinear integral equationswhich involve the use of measure of noncompactness, as well as other techniques, see,for example [1, 2, 6–8,11–17].
After introducing the concept of coupled fixed point by Bhaskar and Lakshmikan-tham [10], many authors used it to generalize the Banach contraction principle, forexample see [3–5, 9, 10, 18, 19]. Recently Aghajani and Sabzali [2] presented somegeneralizations of Darbo’s theorem by using the concept of coupled fixed points.
In this paper, by using the concept of tripled fixed point which was first introducedby [9], we try to present some generalizations of Darbo’s fixed point theorem. The
2010 Mathematics Subject Classification: 47H09, 47H10, 34A12
Keywords and phrases: Measure of noncompactness; system of integral equations; tripledfixed point.
17
18 An existence theorem of tripled fixed point
organization of this paper is as follows. In Section 2, we present notations, definitions,and basic theorems along with some examples. Section 3 is devoted to state and provesome existence theorems on tripled fixed points for a class of operators. In Section 4,using the obtained results in the previous sections, we survey the problem of existenceof solution for the system of nonlinear integral equations (2) and at the end of thissection, there are two examples to illustrate the obtained results.
2. Preliminaries
Let us recall some definitions, notations and preliminary results from the theory ofmeasure of noncompactness in Banach spaces.
From now on, let (E, ‖·‖) be a real Banach space with zero element 0 and Brdenote the closed ball in E centered at 0 with radius r. Denote by ME the family ofnonempty bounded subsets of E and by NE its subfamily consisting of all relativelycompact subsets of E. If X is a subset, we assume that X, co(X) are the closure andclosed convex hull of X in E, respectively.
Definition 2.1 ( [8]). A mapping µ : ME → [0,∞) is said to be a measure ofnoncompactness in E if it satisfies the following conditions:(MNC1) The family kerµ = X ∈ME : µ(X) = 0 is nonempty and kerµ ⊆ NE .
(MNC2) X ⊂ Y ⇒ µ(X) ≤ µ(Y ).
(MNC3) µ(X) = µ(X).
(MNC4) µ(coX) = µ(X).
(MNC5) µ(λX + (1− λ)Y ) ≤ λµ(X) + (1− λ)µ(Y ) for λ ∈ [0, 1].
(MNC6) If (Xn) is a sequence of closed sets from ME such that Xn+1 ⊂ Xn,n = 1, 2, . . . and if limn→∞ µ(Xn) = 0, then the intersection set X∞ =
⋂∞n=1Xn
is nonempty.
The family kerµ described in (MNC1) is said to be the kernel of the measure ofnoncompactness µ. Observe that the intersection set X∞ from (MNC6) is a memberof the family kerµ. In fact, since µ(X∞) ≤ µ(Xn) for any n, we infer that µ(X∞) = 0.This yields that X∞ ∈ kerµ.
Definition 2.2 ( [9]). An element (x, y, z) ∈ X × X × X is called a tripled fixedpoint of a mapping T : X × X × X → X if T (x, y, z) = x and T (y, x, y) = y andT (z, y, x) = z.
Theorem 2.3 ([8]). Suppose that µ1, µ2, . . . , µn are measures of noncompactness inE1, E2, . . . , En, respectively. Moreover, assume that a function F : [0,∞)n → [0,∞)is convex and F (x1, x2, . . . , xn) = 0 if and only if xi = 0 for i = 1, 2, . . . , n. Then
µ(X) = F (µ1(X1), µ2(X2), . . . , µn(Xn))
defines a measure of noncompactness in E1 × E2 × . . . × En, where Xi denotes thenatural projection of X into Ei for i = 1, 2, . . . , n.
B. Matani, J. R. Roshan 19
Now, as illustrations of Theorem 2.3, we present the following examples.
Example 2.4. Let µ be a measure of noncompactness in E. We define F (x, y, z) = x+y+z for any (x, y, z) ∈ [0,∞)3. Then F has the properties mentioned in Theorem 2.3.Hence µ(X) = µ(X1) + µ(X2) + µ(X3) is a measure of noncompactness in the spaceE × E ×E where Xi (i = 1, 2, 3) denotes the natural projections of X into E.
Example 2.5. Let µ be a measure of noncompactness in E. We define F (x, y, z) =max x, y, z for any (x, y, z) ∈ [0,∞)3. Then F has the properties mentioned in The-orem 2.3. Hence µ(X) = max µ(X1), µ(X2), µ(X3) is a measure of noncompactnessin the space E × E ×E where Xi (i = 1, 2, 3) denotes the natural projections of Xinto E.
Theorem 2.6 (Darbo [6]). Let Ω be a nonempty, bounded, closed, and convex subsetof a Banach space E and let T : Ω→ Ω be a continuous mapping. Assume that thereexists a constant k ∈ [0, 1) such that µ(T (X)) ≤ kµ(X) for any nonempty X ⊂ Ω.Then T has a fixed point.
In the sequel, let us recall basic notations which were introduced and discussedin [8]. Consider the space of bounded continuous functions BC(R+) with the norm
‖x‖ = sup |x(t)| : t ≥ 0for any x ∈ BC(R+). Moreover, choose a nonempty bounded subset X of BC(R+)and a positive number L > 0. For x ∈ X, ε ≥ 0 and t ∈ R+. Let
ωL(x, ε) = sup|x(t)− x(s)| : t, s ∈ [0, L] , |t− s| ≤ ε,ωL(X, ε) = sup
ωL(x, ε) : x ∈ X
,
ωL0 (X) = limε→0
ωL(X, ε), ω0(X) = limL→∞
ωL0 (X),
X(t) = x(t) : x ∈ X , diamX(t) = sup |x(t)− y(t)| : x, y ∈ Xand consider the measure of noncompactness
µ(X) = ω0(X) + lim supt→∞
diamX(t), (1)
where diamX(t) = sup |x(t)− y(t)| : x, y ∈ X.It can be shown (see [3, 8]) that the function µ(X) defines a measure of noncom-
pactness in the sense of the above accepted definition.
In this paper, we present and prove some existence theorems of tripled fixed pointfor a class of operators. Moreover, as an application, we study the problem of existenceof solutions for the following system of nonlinear integral equations of the form
x(t)=A(t)+f(t, x(ξ(t)), y(ξ(t)), z(ξ(t)))ϕ(∫ β(t)0
g(t, s, x(η(s)), y(η(s)), z(η(s)))ds),
y(t)=A(t)+f(t, y(ξ(t)), x(ξ(t)), y(ξ(t)))ϕ(∫ β(t)0
g(t, s, y(η(s)), x(η(s)), y(η(s)))ds),
z(t)=A(t)+f(t, z(ξ(t)), y(ξ(t)), x(ξ(t)))ϕ(∫ β(t)0
g(t, s, z(η(s)), y(η(s)), x(η(s)))ds),
(2)
where A, f, g, ϕ, ξ, η and β satisfy certain conditions.
20 An existence theorem of tripled fixed point
3. Main results
In this section, we present and prove some theorems for the existence of tripled fixedpoint for a special class of operators. This basic result will be used in Section 4.
Theorem 3.1. Let µ be an arbitrary measure of noncompactness in E and Ω be anonempty bounded subset of E. Moreover, assume that T : Ω × Ω × Ω → Ω is acontinuous function such that there exist nonnegative constants k1, k2, k3 with k1 +2k2 + k3 < 1 such that
µ(T (X1 ×X2 ×X3) ≤ k1µ(X1) + k2µ(X2) + k3µ(X3) (3)
for all X1, X2, X3 ⊂ Ω. Then T has a tripled fixed point.
Proof. Define T on Ω × Ω × Ω by the formula T (x, y, z) = (T (x, y, z), T (y, x, y),
T (z, y, x)), for all (x, y, z) ∈ Ω × Ω × Ω. It is easy to see that T is continuous
on Ω × Ω × Ω. Now we show that for any X ⊂ Ω × Ω × Ω, we have µ(T (X)) ≤(k1 + 2k2 + k3)µ(X) where µ is defined by Example 2.4. To obtain our purpose, wetake an arbitrary nonempty subset X of Ω×Ω×Ω. Then by (MNC2) and (3), we get
µ(T (X)) ≤ µ(T (X1 ×X2 ×X3), T (X2 ×X1 ×X2), T (X3 ×X2 ×X1))
= µ(T (X1 ×X2 ×X3)) + µ(T (X2 ×X1 ×X2)) + µ(T (X3 ×X2 ×X1))
≤ k1µ(X1) + k2µ(X2) + k3µ(X3) + k1µ(X2) + k2µ(X1) + k3µ(X2)
+ k1µ(X3) + k2µ(X2) + k3µ(X1)
= (k1 + k2 + k3)µ(X1) + (k1 + 2k2 + k3)µ(X2) + (k1 + k3)µ(X3)
≤ (k1 + 2k2 + k3)µ(X1) + (k1 + 2k2 + k3)µ(X2) + (k1 + 2k2 + k3)µ(X3)
= (k1 + 2k2 + k3)(µ(X1) + µ(X2) + µ(X3)) = (k1 + 2k2 + k3)µ(X).
Hence, by Theorem 2.6, T has a tripled fixed point.
By using the above result, we have the following corollary.
Corollary 3.2. Assume that T : Ω×Ω×Ω→ Ω is a continuous function such thatµ(T (X1 ×X2 ×X3)) ≤ k
3 (µ(X1) + µ(X2) + µ(X3)) for each X1, X2, X3 ⊂ Ω where0 ≤ k < 1 is a constant. Then T has a tripled fixed point.
Proof. Taking k1 = k3 = k3 and k2 = k
6 in Theorem 3.1, we obtain the result.
Theorem 3.3. Let T : Ω× Ω× Ω→ Ω be a continuous function such that
µ(T (X1 ×X2 ×X3) ≤ kmax µ(X1), µ(X2), µ(X3 (4)
for any X1, X2, X3 ⊂ Ω, where µ is an arbitrary measure of noncompactness and k isa constant with 0 ≤ k < 1. Then T has at least a tripled fixed point.
Proof. First note that Example 2.5 implies that µ(X) = max µ(X1), µ(X2), µ(X3)is a measure of noncompactness in the space E×E ×E where Xi (i = 1, 2, 3) denotes
the natural projections of X into E. Also the map T : Ω×Ω×Ω→ Ω×Ω×Ω, whereT (x, y, z) = (T (x, y, z), T (y, x, y), T (z, y, x)) is clearly continuous on Ω×Ω×Ω by its
B. Matani, J. R. Roshan 21
definition. Now we claim that T satisfies all the conditions of Theorem 2.6. To provethis, let X ⊂ Ω×Ω×Ω be a nonempty subset. Then by (MNC2) and (4), we get
µ(T (X)) ≤ µ(T (X1 ×X2 ×X3), T (X2 ×X1 ×X2), T (X3 ×X2 ×X1))
= max µ(T (X1 ×X2 ×X3)), µ(T (X2 ×X1 ×X2)), µ(T (X3 ×X2 ×X1))
≤ max
kmax µ(X1), µ(X2), µ(X3) , kmax µ(X2), µ(X1), µ(X2) ,
kmax µ(X3), µ(X2), µ(X1)
= kmax µ(X1), µ(X2), µ(X3) .
Hence µ(T (X)) ≤ kµ(X). Thus, our conclusion follows from Theorem 2.6.
Corollary 3.4. Let T : Ω× Ω× Ω→ Ω be a continuous function such that
‖T (x, y, z)− T (u, v, w)‖ ≤ kmax ‖x− u‖ , ‖y − v‖ , ‖z − w‖for any (x, y, z), (u, v, w) ∈ Ω× Ω× Ω where 0 ≤ k < 1 is a constant. Then T has atripled fixed point.
Proof. It is easy to see that the map µ : ME → [0,∞) defined by µ(X) = diam(X) is ameasure of noncompactness. Therefore, it is sufficient to prove that the inequality (4)is satisfied. To do this, let X1, X2, X3 ⊂ Ω and (x, y, z), (u, v, w) ∈ X1 × X2 × X3.Then, we get
‖T (x, y, z)− T (u, v, w)‖ ≤kmax ‖x− u‖ , ‖y − v‖ , ‖z − w‖≤kmax diam(X1),diam(X2),diam(X3) .
Thus diam(T (X1 × X2 × X3)) ≤ kmax diam(X1),diam(X2),diam(X3). So, byTheorem 3.3, T has a tripled fixed point.
4. Applications and examples
In this section, we present two applications of Theorem 3.1 which can be used toprove the existence of solution for systems of nonlinear integral equations (2).
Theorem 4.1. Let the following hold.(i) The function A : R+ → R is continuous and bounded.
(ii) The function f : R+ × R× R× R→ R is continuous and there exist k1, k2, k3 ∈[0, 1) such that |f(t, x, y, z)− f(t, u, v, w)| ≤ k1 |x− u|+k2 |y − v|+k3 |z − w| for anyt ≥ 0 and for all x, y, z, u, v, w ∈ R.
(iii) The function defined by t 7→ |f(t, 0, 0, 0)| is bounded on R+.
(iv) The functions ξ, η, β : R+ → R+ are continuous and ξ(t)→∞ as t→∞.
(v) The function ϕ : R+ → R is continuous and there exist positive constants α, δsuch that |ϕ(t1)− ϕ(t2)| ≤ δ |t1 − t2|α for any t1, t2 ∈ R+.
(vi) The function g : R+ × R+ × R× R× R→ R is a continuous function such that
limt→∞
∫ β(t)
0
|g(t, s, x(η(s)), y(η(s)), z(η(s)))−g(t, s, u(η(s)), v(η(s)), w(η(s))| ds = 0 (5)
22 An existence theorem of tripled fixed point
uniformly with respect to x, y, z, u, v, w ∈ BC(R+). In addition, M2(k1 +k2 +k3) < 1where
M2= sup∣∣ϕ(∫ β(t)
0
g(t, s, x(η(s)), y(η(s)), z(η(s)))ds∣∣ : t ∈ R+, x, y, z ∈ BC(R+)
. (6)
Then the equation (2) has at least one solution in BC(R+)×BC(R+)×BC(R+).
Proof. First we define an operator T : BC(R+)×BC(R+)×BC(R+)→ BC(R+) by
T (x, y, z)(t) = A(t)
+ f(t, x(ξ(t)), y(ξ(t)), z(ξ(t)))ϕ(∫ β(t)
0
g(t, s, x(η(s), y(η(s)), z(η(s)))ds).
(7)
Moreover, the space BC(R+)×BC(R+)×BC(R+) is equipped with the norm‖(x, y, z)‖ = ‖x‖+ ‖y‖+ ‖z‖ for any (x, y, z) ∈ BC(R+)×BC(R+)×BC(R+).
Notice that the continuity of T (x, y, z) for any (x, y, z) ∈ BC(R+)×BC(R+)×BC(R+)is obvious. Moreover by (6), (7), (i), (ii) and the triangle inequality, we know that
|T (x, y, z)(t)| ≤ |A(t)|+ [|f(t, x(ξ(t)), y(ξ(t)), z(ξ(t)))− f(t, 0, 0, 0)|+ |f(t, 0, 0, 0)|]
× ϕ(
∫ β(t)
0
|g(t, s, x(η(s), y(η(s)), z(η(s)))| ds)
≤M0 + [k1 |x(ξ(t))|+ k2 |y(ξ(t))|+ k3 |z(ξ(t))|+M1]M2 ≤ r (8)
where M0 = supt∈R+|A(t)|, M1 = supt∈R+
|f(t, 0, 0, 0)| and r = M0+M1M2
1−(k1+k2+k3)M2.
Thus T is well defined and the estimate (8) implies that T maps Br into itself. Now,we prove that T is continuous on Br. For this, take (x, y, z) ∈ Br×Br×Br and ε > 0arbitrarily. Moreover, consider (u, v, w) ∈ Br×Br×Br with
‖(x, y, z)− (u, v, w)‖BC(R+)×BC(R+)×BC(R+) ≤ ε.Then we have
|T (x, y, z)(t)−T (u, v, w)(t)|
≤[∣∣∣∣ f(t, x(ξ(t)), y(ξ(t)), z(ξ(t)))−f(t, u(ξ(t)), v(ξ(t)), w(ξ(t)))
∣∣∣∣]∣∣∣∣∣ϕ(
∫ β(t)
0
g(t, s, x(η(s)), y(η(s)), z(η(s)))ds)
∣∣∣∣∣+ [|f(t, u(ξ(t)), v(ξ(t)), w(ξ(t)))−f(t, 0, 0, 0)|
+ |f(t, 0, 0, 0)|]
∣∣∣∣∣ ϕ(∫ β(t)0
g(t, s, x(η(s)), y(η(s)), z(η(s)))ds)
−ϕ(∫ β(t)0
g(t, s, u(η(s), v(η(s)), w(η(s)))ds)
∣∣∣∣∣≤ [k1 |x(ξ(t))−u(ξ(t))|+k2 |y(ξ(t))−v(ξ(t))|+k3 |z(ξ(t))−w(ξ(t))|]M2
+ [k1 |u(ξ(t))|+k2 |v(ξ(t))|+k3 |w(ξ(t))|+M1] · δ∣∣∣∣ ∫ β(t)0
g(t, s, x(η(s)), y(η(s)), z(η(s)))−g(t, s, u(η(s)), v(η(s)), w(η(s)))ds
∣∣∣∣α≤ (k1+k2+k3)εM2
+ [(k1+k2+k3)r+M1] · δ∣∣∣∣ ∫ β(t)0
g(t, s, x(η(s)), y(η(s)), z(η(s)))−g(t, s, u(η(s)), v(η(s)), w(η(s)))ds
∣∣∣∣α . (9)
B. Matani, J. R. Roshan 23
In addition, from (5), there exists L > 0 such that, if t > L, then∫ β(t)
0
|g(t, s, x(η(s)), y(η(s)), z(η(s)))−g(t, s, u(η(s)), v(η(s)), w(η(s)))| ds ≤(εδ
)α(10)
for any x, y, z, u, v, w ∈ BC(R+). There are now two cases:
Case (a) If t > L, then from (9) and (10), we get
|T (x, y, z)(t)− T (u, v, w)(t)| ≤ [(k1 + k2 + k3)(M2 + r) +M1]ε. (11)
Case (b) If t ∈ [0, L], then using the continuity of g on [0, L] × [0, βL] × [−r, r] ×[−r, r]× [−r, r] we have ωLr (g, ε)→ 0, as ε→ 0.
Thus by the argument similar to those given in (9), we have
|T (x, y, z)(t)− T (u, v, w)(t)|≤ (k1 + k2 + k3) εM2 + [(k1 + k2 + k3) r +M1]
×δ
∣∣∣∣∣∫ β(t)
0
(g (t, s, x (η (s)) , y (η (s)) , z (η (s)))−g (t, s, u (η (s)) , v (η (s)) , w (η (s)))) ds
∣∣∣∣∣α
< (k1 + k2 + k3) εM2 + [(k1 + k2 + k3) r +M1] · δ(βLω
Lr (g, ε)
)α< [(k1 + k2 + k3) (M2 + r) +M1] ε, (12)
where ωLr (g, ε) = sup |g(t1, s, x, y, z)− g(t2, s, x, y, z)| : t1, t2 ∈ [0, L] , |t1 − t2| ≤ ε,s ∈ [0, βL] , x, y, z ∈ [−r, r] and βL = supβ(t) : t ∈ [0, L]. Hence, the inequali-ties (11), (12), imply that T is a continuous function from Br ×Br ×Br into Br.
Now, we only need to show that T satisfies the conditions of Theorem 2.6. Toprove that, let L, ε ∈ R+ and X1, X2 , X3 are arbitrary nonempty subsets of Br andtake t1, t2 ∈ [0, L], such that |t1 − t2| < ε.
Without loss of generality, we may assume that β(t1) < β(t2). We also assumethat (x, y, z) ∈ X1 ×X2 ×X3. Then we get
|T (x, y, z)(t1)−T (x, y, z)(t2)|
≤ |A(t1)−A(t2)|+[∣∣∣∣ f(t1, x(ξ(t1)), y(ξ(t1)), ), z(ξ(t1)))−f(t2, x(ξ(t1)), y(ξ(t1)), z(ξ(t1)))
∣∣∣∣+
∣∣∣∣ f(t2, x(ξ(t1)), y(ξ(t1)), z(ξ(t1)))−f(t2, x(ξ(t2)), y(ξ(t2)), z(ξ(t2)))
∣∣∣∣]∣∣∣ϕ(
∫ β(t)
0
g(t, s, x(η(s), y(η(s)), z(η(s)))ds)∣∣∣
+[|f(t2, x(ξ(t2)), y(ξ(t2)), z(ξ(t2)))−f(t2, 0, 0, 0)|+ |f(t2, 0, 0, 0)|]
×
∣∣∣∣∣ ϕ(∫ β(t1)0
g(t1, s, x(η(s)), y(η(s)),z(η(s)))ds)
−ϕ(
∫ β(t2)
0
g(t2, s, x(η(s)), y(η(s)), z(η(s)))ds)
∣∣∣∣∣≤ |A(t1)−A(t2)|+[ωLr (f, ε)+k1 |x(ξ(t1))−x(ξ(t2))|
+k2 |y(ξ(t1))−y(ξ(t2))|+k3 |z(ξ(t1))−z(ξ(t2))|]M2
+[k1 |x(ξ(t2)|+k2 |y(ξ(t2)|+k3 |z(ξ(t2))|+M1]
× δ
∣∣∣∣∣∫ β(t1)0
g(t1, s, x(η(s)), y(η(s)), z(η(s)))ds
−∫ β(t2)0
g(t2, s, x(η(s)), y(η(s)), z(η(s)))ds
∣∣∣∣∣α
24 An existence theorem of tripled fixed point
≤ωL(A, ε))+[ωLr (f, ε)+k1(ωL(x, ωL(ξ, ε)))+k2(ωL(y, ωL(ξ, ε))+k3(ωL(z, ωL(ξ, ε))]M2
+[(k1+k2+k3)r+M1] · δ(∫ β(t1)
0
∣∣∣∣ g(t1, s, x(η(s)), y(η(s)), z(η(s)))−g(t2, s, x(η(s)), y(η(s)), z(η(s)))
∣∣∣∣ ds+
∫ β(t2)
β(t1)
|g(t2, s, x(η(s)), y(η(s)), z(η(s)))| ds)α
≤ωL(A, ε)+[ωLr (f, ε)+k1(ωL(x, ωL(ξ, ε))+k2(ωL(y, ωL(ξ, ε))+k3(ωL(z, ωL(ξ, ε))]M2
+[(k1+k2+k3)r+M1] · δ(βLωLr (g, ε)+GLr ωL(β, ε))α (13)
where
ωL(ξ, ε) = sup|ξ(t1)− ξ(t2)| : t1, t2 ∈ [0, L] , |t1 − t2| ≤ ε,ωL(β, ε) = sup|β(t1)− β(t2)| : t1, t2 ∈ [0, L] , |t1 − t2| ≤ ε,
ωL(x, ωL(ξ, ε)) = sup|x(t1)− x(t2)| : t1, t2 ∈ [0, L] , |t1 − t2| ≤ ωL(ξ, ε),
ωLr (f, ε) = sup
|f(t1, x, y)− f(t2, x, y)| : t1, t2 ∈ [0, L] , |t1 − t2| ≤ ε,
x, y, z ∈ [−r, r]
,
ωβLr (g, ε) = sup
|g(t1, s, x, y, z)− g(t2, s, x, y, z)| : t1, t2 ∈ [0, L] ,|t1 − t2| ≤ ε,s ∈ [0, βL] , x, y, z ∈ [−r, r]
and GLr = sup|g(t, s, x, y, z)| : t ∈ [0, L] , s ∈ [0, βL] and x, y, z ∈ [−r, r].Since (x, y, z) is an arbitrary element of X1 ×X2 ×X3 in (13), we obtain
ωL(T (X1×X2×X3), ε) ≤ ωL(A, ε))+
ωLr (f, ε)k1ω
L(X1, ω
L (ξ, ε))
+k2ωL(X2, ω
L (ξ, ε))
k3ωL(X3, ω
L (ξ, ε))
M2
+[(k1+k2+k3)r+M1] · δ(βLωLr (g, ε)+GLr ωL(β, ε))α. (14)
Otherwise, by the uniform continuity of f , g on [0, L] × [0, βL] × [−r, r] × [−r, r],[0, L]×[0, βL]×[−r, r]×[−r, r]×[−r, r], respectively, we have ωLr (f, ε)→ 0, ωβL
r (g, ε)→0. Also because of the uniform continuity of ξ, β and A on [0, L], we derive thatωL(ξ, ε) → 0, ωβL
r (β, ε) and ωLr (A, ε) → 0 as ε → 0. But GLr is finite, so taking thelimit from (14) as ε→ 0, we get
ωL0 (T (X1 ×X2 ×X3)) ≤ (k1ωL0 (X1) + k2ω
L0 (X2) + k3ω
L0 (X3))M2. (15)
By letting L→∞ in (15), we obtain
ω0(T (X1 ×X2 ×X3)) ≤ k1M2ω0(X1) + k2M2ω0(X2) + k3M2ω0(X3). (16)
In addition, for arbitrary (x, y, z), (u, v, w) ∈ X1 ×X2 ×X3 and t ∈ R+ we have
|T (x, y, z)(t)− T (u, v, w)(t)|
≤(∣∣∣∣ f(t, x(ξ(t)), y(ξ(t)), z(η(s)))−f(t, u(ξ(t)), v(ξ(t)), w(η(s)))
∣∣∣∣)∣∣∣ϕ(
∫ β(t)
0
g(t, s, x(η(s), y(η(s)), z(η(s)))ds)∣∣∣
+ [|f(t, u(ξ(t)), v(ξ(t)), w(ξ(t)))− f(t, 0, 0, 0)|+ |f(t, 0, 0, 0)|]
×
∣∣∣∣∣ ϕ(∫ β(t)0
g(t, s, x(η(s), y(η(s)), z(η(s)))ds)
−ϕ(∫ β(t)0
g(t, s, u(η(s), v(η(s)), w(η(s)))ds)
∣∣∣∣∣
B. Matani, J. R. Roshan 25
≤(k1 |x(ξ(t))− u(ξ(t))|+ k2 |y(ξ(t))− v(ξ(t))|+ k3 |z(ξ(t)− w(ξ(t))|)M2
+ [k1 |u(ξ(t))|+ k2 |v(ξ(t))|+ k3 |w(ξ(t))|+M1]
× δ∣∣∣∣ ∫ β(t)0
g(t, s, x(η(s)), y(η(s)), z(η(s)))−g(t, s, u(η(s)), v(η(s)), w(η(s)))ds
∣∣∣∣α≤(k1 diamX1(ξ(t) + k2 diamX2(ξ(t)) + k3 diamX3(ξ(t)))M2
+[((k1 + k2 + k3)r +M1)] · δ(∫ β(t)
0
∣∣∣∣ g(t, s, x(η(s)), y(η(s)), z(η(s))−g(t, s, u(η(s)), v(η(s)), w(η(s)))
∣∣∣∣ ds)α (17)
Since (x, y, z), (u, v, w) and t are arbitrary in (17), we conclude that
diamT (X1×X2×X3)(t) ≤ (k1 diamX1(ξ(t))+k2 diamX2(ξ(t)+k3 diamX3(ξ(t)))M2
+[((k1+k2+k3)r+M1)] · δ
∣∣∣∣∣∫ β(t)
0
(g (t, s, x (η (s)) , y (η (s)) , z (η (s)))−g (t, s, u (η (s)) , v (η (s)) , w (η (s)))
)ds
∣∣∣∣∣α
(18)
Suppose that t→∞ in the inequality (18), then using (5) we deduce that
lim supt→∞
diamT (X1 ×X2 ×X3)(t) ≤ (k1 lim supt→∞
diamX1(ξ(t))
+ k2 lim supt→∞
diamX2(ξ(t)) + k3 lim supt→∞
diamX3(ξ(t)))M2. (19)
Now, combining (16) and (19) we obtain
ω0(T (X1 ×X2 ×X3)) + lim supt→∞
diamT (X1 ×X2 ×X3)(t)
≤(k1M2ω0(X1) + k2M2ω0(X2) + k3M2ω0(X3)) + k1M2 lim supt→∞
diamX1(ξ(t))
+ k2M2 lim supt→∞
diamX2(ξ(t)) + k3M2 lim supt→∞
diamX3(ξ(t))
=k1M2(ω0(X1)) + lim supt→∞
diamX1(ξ(t))) + k2M2(ω0(X2))
+ lim supt→∞
diamX2(ξ(t))) + k3M2(ω0(X3) + lim supt→∞
diamX3(ξ(t))). (20)
Therefore, from (1) and (20), we derive that µ(T (X1 × X2 × X3)) ≤ k1M2µ(X1) +k2M2µ(X2)+k3M2µ(X3). Consequently µ(T (X1×X2×X3)) ≤ k′1µ(X1)+k
′
2µ(X2)+k′
3µ(X3), where k′
1 = k1M1, k′
2 = k2M2, k′
3 = k3M2 and k′
1 + k′
2 + k′
3 < 1. Thusby Theorem 3.1, T has a tripled fixed point in BC(R+)×BC(R+)×BC(R+). Thiscompletes the proof.
The following corollary is a special case of Theorem 4.1.
Corollary 4.2. Let the conditions (i)–(iv) of Theorem 4.1 be satisfied. Furthermore,suppose that there exists continuous functions a, b : R+ → R+ such that
|g(t, s, x, y, z)| ≤ a(t)b(s) (21)
for t, s ∈ R+. Also, assume that
limt→∞
a(t)
∫ β(t)
0
b(s) ds = 0 (22)
26 An existence theorem of tripled fixed point
and M′
2(k1 + k2 + k3) < 1 where
M′
2 = supa(t)
∫ β(t)
0
b(s) ds : t ∈ R+
. (23)
Then the system of integral equationsx(t) = A(t) + f(t, x(ξ(t)), y(ξ(t)), z(ξ(t)))(
∫ β(t)0
g(t, s, x(η(s)), y(η(s)), z(η(s))) ds)
y(t) = A(t) + f(t, y(ξ(t)), x(ξ(t)), y(ξ(t)))(∫ β(t)0
g(t, s, y(η(s)), x(η(s)), y(η(s))) ds)
z(t) = A(t) + f(t, z(ξ(t)), y(ξ(t)), x(ξ(t)))(∫ β(t)0
g(t, s, z(η(s)), y(η(s)), x(η(s))) ds)
(24)
has at least one solution in the space BC(R+)×BC(R+)×BC(R+).
Proof. Let us consider ϕ(x) = x in Theorem 4.1. It follows that ϕ is a Lipschitzfunction with constant 1. Now from (23) and inequality (21), we get
M2 = sup∣∣∣ϕ(
∫ β(t)
0
g(t, s, x(η(s)), y(η(s)), z(η(s))) ds)∣∣∣ : t ∈ R+, x, y, z ∈ BC(R+)
= sup∣∣∣∫ β(t)
0
g(t, s, x(η(s)), y(η(s)), z(η(s))) ds∣∣∣ : t ∈ R+, x, y, z ∈ BC(R+)
≤ supa(t)
∫ β(t)
0
b(s) ds : t ∈ R+ = M′
2.
Hence, the above inequality implies that M2(k1 + k2 + k3) < 1. On the other handby (21), (22) and the triangle inequality we have
limt→∞
∫ β(t)
0
|g(t, s, x(η(s)), y(η(s)), z(η(s)))− g(t, s, u(η(s)), v(η(s)), w(η(s)))| ds
≤ 2 limt→∞
a(t)
∫ β(t)
0
b(s) ds = 0
uniformly respect to x, y, z, u, v, w ∈ BC(R+). Thus all of the conditions of Theo-rem 3.1 are satisfied. Therefore, applying Theorem 3.1, we conclude that the equa-tion (24) has at least one solution in the space BC(R+)×BC(R+)×BC(R+).
We finish this section with two examples.
Example 4.3. Let m > 2 and n > 1. We set
x(t) = tt+1 + [ 14 sin(x(3t)) + 2
4 cos(y(3t)) + 14ez(3t)]
×∫ t20
ln(1+sm−2
2 |sinn(x(√s))|)|y(√s)||z(√s)|
(1+sin2n(x(√s)))(1+t2m)(1+y2(
√s))(1+z2(
√s))ds
y(t) = tt+1 + [ 14 sin(y(3t)) + 2
4 cos(x(3t)) + 14ey(3t)]
×∫ t20
ln(1+sm−2
2 |sinn(y(√s))|)|x(√s)||y(√s)|
(1+sin2n(y(√s)))(1+t2m)(1+x2(
√s)(1+y2(
√s)))
ds
z(t) = tt+1 + [ 14 sin(z(3t)) + 2
4 cos(y(3t)) + 14ex(3t)]
×∫ t20
ln(1+sm−2
2 |sinn(z(√s))|)|y(√s)||x(√s)|
(1+sin2n(z(√s)))(1+t2m)(1+y2(
√s))(1+x2(
√s))ds
(25)
Then, we have
f(t, x, y, z) =t
t+ 1+ [
1
4sin(x) +
2
4cos(y) +
1
4ez],
B. Matani, J. R. Roshan 27
g(t, s, x, y, z) =ln(1 + s
m−22 |sinn(x)|) |y| |z|
(1 + sin2n(x))(1 + t2m)(1 + y2)(1 + z2),
A(t) =t
t+ 1, ξ(t) = 3t, η(s) =
√s, β(t) = t2
comparing (25) with (24). Now we survey the conditions of Corollary 4.2:
(i) is clear.
(ii) Assume that t ≥ 0 and x, y, z, u, v, w ∈ R. Then we get
|f(t, x, y, z)−f(t, u, v, w)| ≤ 1
4|sin(x)− sin(u)|+2
4|cos(y)− cos(v)|+1
4|ez−ew|
≤ 1
4|x−u|+2
4|y−v|+1
4|z−w|
for all t > 0. Then the condition (ii) is satisfied with k1 = 14 , k2 = 2
4 , k3 = 14 .
(iii) Clearly, the function |f(t, 0, 0, 0)| = tt+1 + 3
4 is bounded.
(iv) Obviously, ξ, η, q are continuous and ξ(t)→∞ as t→∞.
Now suppose that t, s ∈ R+ and x, y, z ∈ R; then
|g(t, s, x, y, z)| =
∣∣∣∣∣ ln(1 + sm−2
2 |sinn(x)|) |y| |z|(1 + sin2n(x))(1 + t2m)(1 + y2)(1 + z2)
∣∣∣∣∣≤ s
m−22 |sinn(x)|) |y| |z|
(1 + sin2n(x))(1 + t2m)(1 + y2)(1 + z2)≤ 1
4
sm−2
2
(1 + t2m)= a(t) · b(s)
for any t, s ∈ R+, and x, y, z ∈ R. Furthermore
limt→∞
a(t)
∫ t2
0
b(s) ds = limt→∞
1
4(1 + t2m)
∫ t2
0
sm−2
2 ds = limt→∞
tm
2m(1 + t2m)= 0
and M′
2 = supt∈R+
a(t)∫ t20b(s) ds ≤ 1
4m . Hence (k1 + k2 + k3)M′
2 ≤ ( 14 + 2
4 + 14 ) 1
4m < 1.
These imply that the assumptions of Corollary 4.2 are satisfied. Therefore, as aresult of Corollary 4.2, we conclude that the system of integral equations (25) has atleast one solution in BC(R+)×BC(R+)×BC(R+).
Example 4.4. Let n > 1. Consider the equation
x(t) = e−t2
+ [ t2π(t2+1)
cosx(2t) + 24π
tt+1
ln(1 + |y(2t)|) + t4π(t4+1)
sin z(2t)]
· arctg(∫ t0
4n ln(1+sn−1|x(√s)|)|cos y(√s)||sin z(√s)|+ns2n−1(1+|x2(√s)|)(1+cos2 y(√s))(1+sin2 z(
√s))
(1+x2(√s))(1+t2n)(1+cos2 y(
√s))(1+sin2 z(
√s))
ds)
y(t) = e−t2
+ [ t2π(t2+1)
cos y(2t) + 24π
ln(1 + |x(2t)|) + t4π(t4+1)
sin y(2t)]
· arctg(∫ t0
4n ln(1+sn−1|y(√s)|)|cos x(√s)||sin y(√s)|+ns2n−1(1+|y2(√s)|)(1+cos2 x(√s))(1+sin2 y(
√s))
(1+y2(√s))(1+t2n)(1+cos2 x(
√s))(1+sin2 y(
√s))
ds)
z(t) = e−t2
+ [ t2π(t2+1)
cos z(2t) + 24π
ln(1 + |y(2t)|) + t4π(t4+1)
sinx(2t)]
· arctg(∫ t0
4n ln(1+sn−1|z(√s)|)|cos y(√s)||sin x(√s)|+ns2n−1(1+|z2(√s)|)(1+cos2 y(√s))(1+sin2 x(
√s))
(1+z2(√s))(1+t2n)(1+cos2 y(
√s))(1+sin2 x(
√s))
ds)
Since the previous equation is a special case of the equation (2), we obtain
f(t, x, y, z) = e−t2
+ [t
2π(t2 + 1)cos(x) +
2
4π
t
t+ 1ln(1 + |y|) +
1
4π
t
t4 + 1sin(z)]
28 An existence theorem of tripled fixed point
g(t, s, x, y, z) =4n ln(1 + sn−1 |x|) |cos y| |sinz|+ ns2n−1(1 + x2)(1 + cos2 y)(1 + sin2 z)
(1 + x2)(1 + t2n)(1 + cos2 y)(1 + sin2 z)
A(t) = e−t2
, ξ(t) = 2t, η(s) =√s, β(t) = t, ϕ(x) = arctg(x).
We will prove that the conditions of Theorem 4.1 are satisfied for this equation. First,in the same way as stated in Example 4.4, (i), (iii) and (iv) are evident. Now, supposethat t ∈ R+ and x, y, z ∈ R with |y| ≥ |v|. Then we get
|f(t, x, y, z)−f(t, u, v, w)| ≤ t
2π(t2+1)|cos(x)− cos(u)|
+2
4π
t
t+1|ln(1+ |y|)− ln(1+ |v|)|+ t
4π(t4+1)|sin(z)−sin(w)|
≤ 1
4π|x−u|+ 2
4π
∣∣∣∣ln(1+ |y|1+ |v|
)
∣∣∣∣+ 1
4π|z−w| ≤ 1
4π|x−u|+ 2
4π
∣∣∣∣ln(1+|y| − |v|1+ |v|
)
∣∣∣∣+ 1
4π|z−w|
≤ 1
4π|x−u|+ 2
4πln(1+ |y−v|)+ 1
4π|z−w| ≤ 1
4π|x−u|+ 2
4π|y−v|+ 1
4π|z−w| .
Therefore f satisfies the condition (ii) of Theorem 4.1 with k1 = 14π , k2 = 2
4πandk3 = 1
4π . On the other hand, according to the Mean Value Theorem, ϕ is Lipschitzwith constant 1. This means that the condition (v) of Theorem 4.1 is satisfied. Also,it is obvious that g is continuous on R+ × R+ × R× R× R. Since
nsn−1 |x| |cos y| |sinz|(1 + x2)(1 + t2n)(1 + cos2 y)(1 + sin2 z)
≤ 1
4
nsn−1
(1 + t2n)
for any t, s ∈ R+ and x, y, z ∈ R, we have
limt→∞
∫ t
0
|g(t, s, x, y, z)| ds = limt→∞
∫ t
0
ns2n−1
(1 + t2n)ds =
1
2.
Notice that this shows that the property (22) of Corollary 4.2 is not satisfied and thiscorollary cannot be used to conclude that the considered equation has a solution. But∫ t
0
|g(t, s, x, y, z)| ds ≤∫ t
0
4nsn−1 |x| |cos y| |sin z|(1 + x2)(1 + t2n)(1 + cos2 y)(1 + sin2 z)
ds+
∫ t
0
ns2n−1
(1 + t2n)ds
≤∫ t
0
nsn−1
(1 + t2n)ds+
∫ t
0
ns2n−1
(1 + t2n)ds ≤ tn
(1 + t2n)+
t2n
2(1 + t2n)≤ 1
2+
1
2= 1
for any t ∈ R+ and x, y, z ∈ R. Thus
M2 = supt∈R+
∣∣∣arctg(
∫ t
0
g(t, s.x, y, z) ds)∣∣∣ ≤ sup
x∈[−1,1]|arctg(x)| = π
4.
It follows that (k1 + k2 + k3)M2 = ( 14π + 2
4π + 14π )π4 < 1. Moreover
limt→∞
∫ t
0
|g(t, s, x, y, z)− g(t, s, u, v, w)| ds ≤ limt→∞
2
∫ t
0
nsn−1
(1 + t2n)ds = lim
t→∞
2tn
(1 + t2n)= 0.
Hence the condition (vi) of Theorem 4.1 is satisfied too. Now, according to Theo-rem 4.1, we have a solution in BC(R+)×BC(R+)×BC(R+).
B. Matani, J. R. Roshan 29
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(received 09.06.2018; in revised form 27.11.2018; available online 30.04.2019)
Department of Mathematics, Qaemshahr Branch, Islamic Azad University, Qaemshahr, Iran
E-mail: [email protected]
Department of Mathematics, Qaemshahr Branch, Islamic Azad University, Qaemshahr, Iran
E-mail: [email protected], [email protected]
